The equation whose roots are the squares of the roots of x3+ax+b=0, is
A
x3+2ax2−a2x−b2=0
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B
x3+2ax2+a2x−b2=0
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C
x6+ax2+6=0
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D
x6−ax2+6=0
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Solution
The correct option is Bx3+2ax2+a2x−b2=0 Let y=x2 x3+ax+b=0⇒x3+ax=−b⇒(x3+ax)2=(−b)2⇒x6+a2x2+2ax4=b2 Substituting y=x2, we get y3+a2y+2ay2=b2⇒y3+2ay2+a2y−b2=0 Now replacing y by x x3+2ax2+a2x−b2=0 is the eqaution whose roots are square of roots of x3+ax+b=0